poj 2049 Finding Nemo (反过来使用bfs)

poj 2049

题意:  Nemo被困在迷宫里， (其坐标为double型) , 迷宫的每个格子的任一面都有两个种可能:  门,墙; 在迷宫之外的为空地, 要求若Nemo的父亲能找到Nemo, 输出其经过最少的门, 否则输出-1；

解决方法:题目是父亲找Nemo, 可以当作Nemo如何经过最少的门走出迷宫, 每次仅移动一格, 所以可用bfs, 以Nemo所在方格为起点遍历迷宫, 每个方格用其左下角的坐标表示; 采用优先队列进行优化, 每次取出当前已经过门最少的位置.

code:

/***************
Problem from :
Problem describe :
data:
****************/

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#include<stack>
#include<queue>
#include<ctime>
#include<cstring>
#include<vector>
#include<string>
#define ll __int64
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 500;
int n, m, ans;
int xb[maxn][maxn], yb[maxn][maxn], dis[maxn][maxn];
struct node{
int x;
int y;
int doors;
bool operator < (const node &a) const {
return doors>a.doors;
}
};
bool judge(int sx, int sy){
if(sx<0 || sy<0) return true;
//该方格为空地, 故已经走出迷宫
if(xb[sx][sy] == -1 || xb[sx][sy+1] ==-1 || yb[sx][sy]==-1 || yb[sx+1][sy]==-1) return true;
else return false;
}
//以左下角坐标表示每一个小格
int bfs(int sx, int sy)
{
//cout << sx <<" "<<sy<<endl;
if(judge(sx, sy)) return 0;
memset(dis, inf, sizeof(dis));
priority_queue<node>que;
int i, cnt;
node p;
p.x = sx;
p.y = sy;
p.doors = 0;
dis[sx][sy] = 0;
que.push(p);
while(!que.empty())
{
node q = que.top();
que.pop();
sx = q.x;
sy = q.y;
cnt = q.doors;
//cout << sx << " " << sy <<endl;
//向上走
if(xb[sx][sy+1] == 0) {
if(judge(sx, sy+1)) return cnt+1;
p.x = sx;
p.y = sy+1;
p.doors = cnt+1;
//确保从起点到达每个格子经过的门都是最少的
if(dis[p.x][p.y] > p.doors) dis[p.x][p.y] = p.doors, que.push(p);
}
//向下走
if(xb[sx][sy] == 0){
if(judge(sx, sy-1)) return cnt+1;
p.x = sx;
p.y = sy-1;
p.doors = cnt+1;
if(dis[p.x][p.y] > p.doors) dis[p.x][p.y] = p.doors, que.push(p);
}
//向左走
if(yb[sx][sy] == 0){
if(judge(sx-1, sy)) return cnt+1;
p.x = sx-1;
p.y = sy;
p.doors = cnt+1;
if(dis[p.x][p.y] > p.doors) dis[p.x][p.y] = p.doors, que.push(p);
}
//向右走
if(yb[sx+1][sy] == 0){
if(judge(sx+1, sy)) return cnt+1;
p.x = sx+1;
p.y = sy;
p.doors = cnt+1;
if(dis[p.x][p.y] > p.doors) dis[p.x][p.y] = p.doors, que.push(p);
}
}
return -1;
}
int main()
{
//freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
int x, y, r, l, i, j;
while(scanf("%d %d", &n, &m), n!=-1 && m!=-1){
memset(xb, -1, sizeof(xb));//-1表示此处非墙也非门, 即该方格为空地
memset(yb, -1, sizeof(yb));
for(i=1; i<= n; i++){
scanf("%d %d %d %d", &x, &y, &r, &l);
if(r==1){ //与y轴平行
for(j=y; j<l+y; j++) { //注意 这里要<号 而不是等于号
yb[x][j] = 1;
}
} else { //与x轴平行
for(j=x; j<l+x; j++) {
xb[j][y] = 1;
}
}
}
for(i=1; i<=m; i++){
scanf("%d %d %d", &x, &y, &r);
if(r==1) {
yb[x][y]=0;
} else {
xb[x][y]=0;
}
}
double s_x, s_y;
scanf("%lf %lf", &s_x, &s_y);
//cout << s_x <<s_y<<endl;
if(s_x > 199 || s_y>199 || s_x<1||s_y<1) printf("%d\n", 0);
else printf("%d\n",bfs(int(s_x), int(s_y)));
}
return 0;
}